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From the uncertainty relation it follows that:

$$\Delta E \ \Delta t = \hbar$$

$\Delta E$ is the energy uncertainty of a state, $\Delta t$ should be the uncertainty of the lifetime $\tau_b$ of the state. I don't understand why $\Delta t$ is usually confounded with the lifetime itself, $\tau_b$ and the relation written as

$$\Delta E \ \tau_b = \hbar$$

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If we describe a system to be in a specific state $E_0$ at time $t=0$, but this state has a finite lifetime $\tau_b$, the state decays in time according to $$ \psi(t)\sim e^{-t/2\tau_b}e^{iE_0t/\hbar}.$$

This state can be written as a superposition of states with a range of energies around $E_0$ according to $$ \psi(t) \sim \frac{1}{2\pi\hbar}\int_{-\infty}^{+\infty}dE \frac{1}{1-2i(E-E_0)\tau_b/\hbar}e^{-i(E-E_0) t/\hbar}. $$ Note that this is essentially the Fourier transform of the exponential decay, if we use the Einstein relation between energy and frequency $E=\hbar\omega$.

The energy-uncertainty of this state, taken to be the FWHM $\Delta E$ of the Lorentzian energy distribution $$ p(E)dE = \frac{dE}{1+(E-E_0)^2(2\tau_b)^2/\hbar^2}$$ obeys the equation $$ (\Delta E/2)^2(2\tau_b)^2/\hbar^2 = 1 \Rightarrow \Delta E\tau_b = \hbar.$$

This shows how the decay time $\tau_b$ of an exponentially decaying state is related (via a Fourier-transform) to the energy uncertainty $\Delta E$ of the state.

I do not consider this explanation to be a rigorous derivation, and I am sure it can be formulated in a more rigorous way. However, I like it, because it exemplifies the relation between 'decay in time' and 'energy uncertainty' in a very transparent way.

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When you say usually, you are 'usually' sweeping a lot of details under the rug. For example, the motivation of taking $\Delta t = \tau_b$ could be to see what the minimal energy uncertainty could be (minimum uncertainty in energy will correspond to maximum uncertainty in time, which cannot be greater than the actual lifetime itself). If this is a situation specific matter, then of course the details, and hence, assumptions will differ. If you could refer an example where you have seen the lifetime equalling uncertainty assumption, the answer will be clearer.

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  • $\begingroup$ It is a key relation in optical spectroscopy also used to determine the natural bandwidth $\endgroup$ – Giuliano Malatesta Feb 15 '19 at 11:20

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